- #1
Sahil Kukreja
- 65
- 15
Homework Statement
Let P(x) be a polynomial of least degree whose graph has three points of inflection (-1,-1) ; (1,1) and a point with abscissa 0 at which the curve is inclined to the axis of abscissa at an angle of ## \frac {\pi}{3} , Then \int_0^1 P(x) \,dx = ? ##
Homework Equations
No equations Relevant.
The Attempt at a Solution
Since P(x) has three inflection points at x=-1,x=0,x=1
=>## P''(x)= a(x-1)(x)(x+1) = a(x^3 - x) ##
=> ## P'(x) = a( \frac {x^4}{4} - \frac {x^2}{2}) + c1 ##
## now~ it~ is ~given ~that~ P'(0) = \sqrt 3 ##
## => c1= \sqrt 3 ##
=> ## P(x)= a( \frac {x^5}{20} - \frac {x^3}{6} ) + \sqrt 3 x +c2 ##
now P(-1) = -1 and P(1)= 1
solving for a and c2
## P(x)= \frac {60(\sqrt 3 - 1)}{7} ( \frac {x^5}{20} - \frac {x^3}{6} ) + \sqrt 3 x ##
## \int_0^1 P(x)dx = \frac {3 \sqrt 3 + 4 }{14} ##
The question is solved but it is a very lengthy method, can there be a quicker method??
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